AS CAIE Statistics: Unit Test Mock Paper Analysis | AS CAIE 统计：单元测试模拟卷解析

📚 AS CAIE Statistics: Unit Test Mock Paper Analysis | AS CAIE 统计：单元测试模拟卷解析

This walkthrough of a mock unit test for AS CAIE Statistics breaks down typical exam-style questions, revealing the logical steps and common pitfalls. By studying these worked solutions, you can strengthen your understanding of data handling, probability, distributions and statistical inference.

本篇 AS CAIE 统计单元模拟卷解析拆解了典型考题，展示了完整的解题逻辑与常见易错点。通过研读这些详细解答，你可以巩固数据处理、概率、分布与统计推断的核心技能。

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1. Stem-and-Leaf Diagrams and Five-Number Summary | 茎叶图与五数概括

The raw data below show the times (in minutes) taken by 15 students to complete a puzzle: 12, 15, 9, 23, 17, 14, 8, 21, 13, 19, 16, 22, 11, 18, 20. Construct a stem-and-leaf diagram and find the median, quartiles and interquartile range.

原始数据记录 15 名学生完成谜题的时间（分钟）：12, 15, 9, 23, 17, 14, 8, 21, 13, 19, 16, 22, 11, 18, 20。要求绘制茎叶图，并求中位数、四分位数和四分位距。

Step 1: Sort the data in ascending order. The ordered list is 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23.

第一步：将数据按升序排列。排序后为 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23。

Step 2: For the stem-and-leaf diagram, use the tens digit as the stem and the units digit as the leaf. The stem 0 will hold leaves 8,9; stem 1 will hold 1,2,3,4,5,6,7,8,9; stem 2 will hold 0,1,2,3. Remember to provide a key, e.g. 1|2 means 12.

第二步：绘制茎叶图时，十位数为茎，个位数为叶。茎 0 对应叶 8,9；茎 1 对应叶 1,2,3,4,5,6,7,8,9；茎 2 对应叶 0,1,2,3。务必写出图例，如 1|2 表示 12。

st 7 values have median at position 4, giving Q₁ = 12. The upper quartile Q₃ is the median of the upper half: position 12 gives Q₃ = 20. The interquartile range IQR = Q₃ − Q₁ = 8.

第三步：确定中位数的位置：(15+1)/2 = 8，第八个数据为中位数，即 15。下四分位数 Q₁ 为前半数据的中位数：前 7 个数的第 4 位，Q₁ = 12。上四分位数 Q₃ 为后半数据的中位数：第 12 位，Q₃ = 20。四分位距 IQR = Q₃ − Q₁ = 8。

1.5×IQR or above Q₃ + 1.5×IQR; here the boundaries are 0 and 32, so there are no outliers.

第四步：五数概括（最小值 8，Q₁=12，中位数 15，Q₃=20，最大值 23）可用于绘制箱线图。离群值通常指小于 Q₁ − 1.5×IQR 或大于 Q₃ + 1.5×IQR 的数据；此处界限为 0 和 32，因此无离群值。

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2. Mean, Variance and Standard Deviation | 均值、方差与标准差

Using the same data set (8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23), calculate the sample mean, sample variance and standard deviation.

使用同一组数据 (8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23)，计算样本均值、样本方差和标准差。

23. Using the arithmetic series formula, sum = (8+23)×16/2 = 31×8 = 248. The sample mean x̄ = 248/15 ≈ 16.533.

第一步：15 个数据总和：8+9+11+12+13+14+15+16+17+18+19+20+21+22+23。利用等差数列求和，总和 = (8+23)×16/2 = 31×8 = 248。样本均值 x̄ = 248/15 ≈ 16.533。

Step 2: Compute the sum of squared deviations Σ(x − x̄ − 4101.067 = 456.933 (rounded). Then sample variance s² = Sxx/(n−1) = 456.933/14 ≈ 32.638.

第二步：计算离差平方和 Σ(x − x̄)²。尽管计算器可求出 Σx² = 8²+9²+...+23² = 4558，我们可利用公式 Sxx = Σx² − (Σx)²/n = 4558 − 248²/15 = 4558 − 4101.067 = 456.933。样本方差 s² = Sxx/(n−1) = 456.933/14 ≈ 32.638。

Step 3: Standard deviation s = √s² ≈ √32.638 ≈ 5.713. Always show your working and round to an appropriate degree of accuracy.

第三步：标准差 s = √s² ≈ √32.638 ≈ 5.713。注意展示计算过程并保留合理精度。

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3. Probability Rules and Venn Diagrams | 概率运算与维恩图

In a class of 40 students, 24 study Mathematics (M), 18 study Physics (P), and 10 study both. Construct a Venn diagram and find the probability that a randomly selected student studies neither subject.

某班有 40 名学生，其中 24 人学习数学 (M)，18 人学习物理 (P)，10 人同时学习两科。绘制维恩图并求随机选取一名学生既不学数学也不学物理的概率。

ly = 24 − 10 = 14, P only = 18 − 10 = 8. The number studying neither = 40 − (14+10+8) = 8.

第一步：将交集 10 填入重叠部分。则只学数学为 24 − 10 = 14，只学物理为 18 − 10 = 8。两科都不学的人数 = 40 − (14+10+8) = 8。

nn diagram clearly shows the four regions and helps us verify that total probabilities sum to 1.

第二步：P(都不学) = 8/40 = 0.2。维恩图清晰展示了四个区域，帮助我们验证总概率之和为 1。

Step 3: Using probability notation, P(M∪P) = P(M) + P(P) − P(M∩P) = (24+18−10)/40 = 32/40 = 0.8. Thus P(neither) = 1 − 0.8 = 0.2, confirming the result.

第三步：利用概率符号，P(M∪P) = P(M) + P(P) − P(M∩P) = (24+18−10)/40 = 32/40 = 0.8。因此 P(都不学) = 1 − 0.8 = 0.2，验证了结果。

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4. Conditional Probability | 条件概率

Using the same class data, find the probability that a student studies Mathematics given that the student studies Physics.

沿用同班数据，求在已知某学生学物理的条件下该生也学数学的概率。

Step 1: Apply the conditional probability formula: P(M|P) = P(M∩P) / P(P).

第一步：应用条件概率公式：P(M|P) = P(M∩P) / P(P)。

Step 2: P(M∩P) = 10/40 = 0.25. P(P) = 18/40 = 0.45. Therefore P(M|P) = 0.25 / 0.45 = 5/9 ≈ 0.556.

第二步：P(M∩P) = 10/40 = 0.25，P(P) = 18/40 = 0.45。因此 P(M|P) = 0.25 / 0.45 = 5/9 ≈ 0.556。

Step 3: Interpret the result: about 55.6% matics. This is different from P(M) = 24/40 = 0.6, showing the events are not independent.

第三步：解读结果：约 55.6% 的学物理学生同时也学数学。这不同于 P(M) = 24/40 = 0.6，表明两事件不独立。

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5. Discrete Random Variables and Expectation | 离散随机变量与期望

y distribution: P(X=1)=0.2, P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.1. Calculate E(X) and Var(X).

离散随机变量 X 的概率分布为：P(X=1)=0.2, P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.1。计算 E(X) 与 Var(X)。

Step 1: E(X) = Σ x·P(X=x) = 1·0.2 + 2·0.3 + 3·0.4 + 4·0.1 = 0.2 + 0.6 + 1.2 + 0.4 = 2.4.

第一步：E(X) = Σ x·P(X=x) = 1·0.2 + 2·0.3 + 3·0.4 + 4·0.1 = 0.2 + 0.6 + 1.2 + 0.4 = 2.4。

Step 2: Compute E(X²) = 1²·0.2 + 2²·0.3 + 3²·0.4 + 4²·0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6.

第二步：计算 E(X²) = 1²·0.2 + 2²·0.3 + 3²·0.4 + 4²·0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6。

6 = 0.84. Alternatively, Var(X) = Σ(x−μ)²P(x), but the shortcut formula is more efficient.

第三步：Var(X) = E(X²) − [E(X)]² = 6.6 − (2.4)² = 6.6 − 5.76 = 0.84。也可利用 Var(X) = Σ(x−μ)²P(x)，但公式捷径更高效。

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6. Binomial Distribution | 二项分布

A factory produces components, and 25% are defective. A batch of 10 components is selected randomly. Using X ~ B(10, 0.25), find the probability that exactly 3 are defective, and the probability that at least 2 are defective.

某工厂生产的零件中有 25% 为次品。随机抽取 10 个零件。设 X ~ B(10, 0.25)，求恰好有 3 个次品的概率，以及至少 2 个次品的概率。

Step 1: For P(X = 3), use the binomial probability formula: P(X = r) = ⁿC_r p^r (1−p)^n−r. Here ¹⁰C₃ ⁷. 0.25³ = 0.015625, 0.75⁷ ≈ 0.13348, product ≈ 120 × 0.002084 ≈ 0.250 (precisely 0.2503).

第一步：P(X = 3) 用二项概率公式：P(X = r) = ⁿC_r p^r (1−p)^n−r。此处 ¹⁰C₃ = 120，所以 P(X=3) = 120 × 0.25³ × 0.75⁷。0.25³ = 0.015625，0.75⁷ ≈ 0.13348，乘积 ≈ 120 × 0.002084 ≈ 0.250（精确值为 0.2503）。

⁹ ≈ 10 × 0.25 × 0.07508 ≈ 0.1877. So P(X ≥ 2) = 1 − (0.0563 + 0.1877) = 1 − 0.244 = 0.756.

第二步：P(X ≥ 2) = 1 − [P(X=0) + P(X=1)]。计算 P(X=0) = 0.75¹⁰ ≈ 0.0563；P(X=1) = 10 × 0.25 × 0.75⁹ ≈ 10 × 0.25 × 0.07508 ≈ 0.1877。因此 P(X ≥ 2) = 1 − (0.0563 + 0.1877) = 1 − 0.244 = 0.756。

Step 3: Always check if using cumulative binomial tables or a calculator; in exams, show the formula and substitution clearly. The results are rounded to three decimal places as appropriate.

第三步：考试中可使用二项分布累计表或计算器，但仍需清晰展示公式及代入过程。结果通常保留三位小数。

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7. Normal Distribution: Finding Probabilities | 正态分布：求概率

The masses of apples from an orchard are normally distributed with mean 150 g and standard deviation 20 g. Find P(mass < 175 g) and P(140 < mass < 165).

某果园苹果质量服从正态分布，均值为 150 g，标准差为 20 g。求 P(质量 < 175 g) 和 P(140 < 质量 < 165)。

Step 1: For X ~ 0 = 1.25. Then P(X < 175) = P(Z < 1.25). From the standard normal table, Φ(1.25) = 0.8944.

第一步：设 X ~ N(150, 20²)，标准化 Z = (X − μ)/σ。对于 175：Z = (175 − 150)/20 = 1.25，则 P(X < 175) = P(Z < 1.25)。查标准正态表得 Φ(1.25) = 0.8944。

.5, Z₂ = (165 − 150)/20 = 0.75. P(140 < X < 165) = P(−0.5 < Z < 0.75) = Φ(0.75) − Φ(−0.5).

第二步：对于区间，计算两个 Z 值：Z₁ = (140 − 150)/20 = −0.5，Z₂ = (165 − 150)/20 = 0.75。P(140 < X < 165) = P(−0.5 < Z < 0.75) = Φ(0.75) − Φ(−0.5)。

robability = 0.7734 − 0.3085 = 0.4649. Always sketch a bell curve to visualise the region.

第三步：Φ(0.75) = 0.7734，Φ(−0.5) = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085。因此概率 = 0.7734 − 0.3085 = 0.4649。建议画出正态曲线草图以直观确认区域。

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8. Normal Distribution: Inverse Normal | 正态分布：逆向查表

Using the same apple distribution N(150, 20²), find the mass k such that 10% of apples weigh more than k.

沿用相同的苹果分布 N(150, 20²)，求质量 k，使得 10% 的苹果质量超过 k。

e Z-value such that Φ(z) = 0.90. From tables, z ≈ 1.2816 (or use inverse normal function).

第一步：P(X > k) = 0.10 等价于 P

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